Photoproduction of {\boldmath{$\pi^{0}$}} Mesons off Protons and Neutrons in the Second and Third Nucleon Resonance Region

The A2 collaboration Dieterle, M. ; Werthmüller, D. ; Abt, S. ; et al.
Phys.Rev.C (2018) 065205-1-065205-28, 2018.
Inspire Record 1675023 DOI 10.17182/hepdata.131794

Photoproduction of mesons off quasi-free nucleons bound in the deuteron allows to study the electromagnetic excitation spectrum of the neutron and the isospin structure of the excitation of nucleon resonances. The database for such reactions is much more sparse than for free proton targets. Single $\pi^0$ photoproduction off quasi-free nucleons from the deuteron was experimentally studied. Nuclear effects were investigated by a comparison of the results for free protons and quasi-free protons and used as a correction for the quasi-free neutron data. The experiment was performed at the tagged photon beam of the Mainz MAMI accelerator for photon energies between 0.45~GeV and 1.4~GeV, using an almost $4\pi$ electromagnetic calorimeter composed of the Crystal Ball and TAPS detectors. A complete kinematic reconstruction of the final state removed the effects of Fermi motion. Reaction model predictions and PWA for $\gamma n\rightarrow n\pi^{0}$, based on fits to data for the other isospin channels, disagreed between themselves and no model provided a good description of the new data. The results demonstrate clearly the importance of a measurement of the fully neutral final state for the isospin decomposition of the cross section. Model refits, for example from the Bonn-Gatchina analysis, show that the new and the previous data for the other three isospin channels can be simultaneously described when the contributions of several partial waves are modified. The results are also relevant for the suppression of the higher resonance bumps in total photoabsorption on nuclei, which are not well understood.

19 data tables

Excitation function at cos(Theta_pi0)cm = -0.95

Excitation function at cos(Theta_pi0)cm = -0.85

Excitation function at cos(Theta_pi0)cm = -0.75

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